doug@zaphod.prime.com (12/30/88)
Ohms Law: V = IZ
2 2
Power Law: E = IV = I R = E / R
First, the current is indeed determined by Ohm's law. The problem is that a
car starter does not have a constant impedance. When the starter is turning
over quickly (low load, as in a warm day) the current draw is only 10->20
amps. On a cold day the starter has to work much harder and will often draw
100 or more amps. Note that all this is directly related to the work
(force x distance) the starter must do. So on a warm day the starter is
using between 120 and 240 watts energy and on a cold day upward of 1.2 Kw.
In general for power reasons it is better to have high voltage and low current
in a line. This is because the primary power loss is due to resistive voltage
drop in the line and is directly proportional to the current. You can beat
this by lowering the current (therefore raising the voltage) or lowering the
resistance (e.g. using superconductors). That's why power companies transmit
power cross country at hundreds of thousands of volts and do local power at
several thousand volts.
Douglas Rand
Internet: doug@zaphod.prime.com
Usenet: primerd!zaphod!doug
Phone: (508) - 879 - 2960
Mail: Prime Computer, 500 Old Conn Path, MS10C-17, Framingham, Ma 01701
-> The above opinions are mine alone.myers@hpfcdj.HP.COM (Bob Myers) (12/30/88)
> Could someone please explain why it is better to have a circuit with >high amps and low volts rather than the other way around? Does increased amps >lead to increased power in the same proportion as increased voltages leads to >increased power? Why are car batteries touted as having more cranking amps, >isn't your amperage held constant by: > V = I R V = 12, R = Constant Power = V^2/R > What am I missing here? A brain perhaps? > Thanks in advance, -Mark The fact that you can ask the question is indication enough that your brain is functioning quite nicely, thank you; one of my biggest problems as an electronics instructor has been to get students to *ASK QUESTIONS!!!!* about anything they don't completely understand. Now, dragging out my standard "Voltage, Current, and Power" lecture..... The basic definition of power is voltage times current, or, in equation form: P = V * I (The power dissipated in a circuit element is equal to the potential across that element (a resistor, say) in volts, multiplied by the current through that element in amperes). Now, since Ohm's Law gives us the relation between voltage, current, and resistance (or, in the more general form, impedance), we can substitute into the above as follows: IF we know voltage and resistance, then I = V/R, so P = (V/R)*V, or P = V^2/R IF we know current and resistance, then V = I * R, so P = (I * R) * I, or P = I^2 * R Either expression is equivalent to the original definition of power, through Ohm's Law. Obviously, then, increasing either the current through a resistance, or the voltage across it, will increase the power which that resistance dissipates. But, equally obviously, increasing either or these also increases the other, by Ohm's Law. They're really two ways of looking at the same thing - there is really no distinction between "increasing the voltage across the resistor" and "increasing the current through a resistor", as Ohm's Law shows that these quantities are always directly related. (One classical analogy is water - electrical current is analogous to gallons per minute, or some other volume/time expression, while electrical potential (voltage) is analogous to water pressure. You can get the same "power" from a stream of water in either of the following cases - a large pipe (high current) with relatively low pressure (voltage), or a small pipe (low current) with high pressure (voltage). The size of the pipe is analogous to the resistance. In theory, then, there is no difference, power-wise, between high voltage/low- current situations, and the other way around. However, in practice, it is much simpler to deal with high currents than to deal with high voltages. High currents simply require bigger conductors. High voltage, on the other hand, can require sophisticated insulation and isolation schemes to "keep the electricity where it belongs", as it wants to keep breaking down (ionizing) such common materials as the surrounding air, and hence "getting away". In addition, high voltage can be extremely hazardous. You've probably heard that it's current that's dangerous, and that's true; but people are pretty high- resistance items, and it takes a fairly high voltage to force a dangerous level of current through dry, unbroken skin. Voltages on the order of the low tens of volts are considered "safe", as they are not high enough to cause appreciable current flow through the body. You don't feel much if you grab the terminals of your car battery by accident, but you'll certainly feel the line current from an AC outlet - and yet the battery is "capable of supplying more current". It's just that it can't supply that current through a high resistance, since it's voltage it too low. Now, what about the "cranking amps" rating of the battery? Well, it's true that all car batteries are 12V (actually, more like 13V) devices; what distinguishes them is their internal resistance. A battery can be modelled as a voltage source (something which produces a constant voltage across its terminals, regardless of the current through it), in series with a resistance: ____/\/\/\/\/\________ + + | Rint ___ | | |12V | Vout ----- | - _______________________ - Note that with no external resistance connected, the voltage Vout is exactly the voltage of the source: 12 volts. However, as soon as any resistance is connected, and current begins to flow, there is a drop across the internal resistance, and the voltage Vout must be something less than the source. If the load resistor is very small, then Rint acts to limit the current which the source supplies. The worst case is, of course, a zero-ohm load - a short across the terminals. In this case, the maximum current (and so the max. current that the battery can EVER produce) is 12V/Rint. All batteries have some internal resistance; it's part of the physical structure of the battery. Obviously, the lower this resistance, the better, at least from the standpoint of our desire to have an "ideal" voltage source. Hope this has helped! Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not Ft. Collins, Colorado | those of my employer or any other {the known universe}!hplabs!hpfcla!myers | sentient life-form on this planet.
ron@hpfcmr.HP.COM (Ron Miller) (12/30/88)
> > Could someone please explain why it is better to have a circuit with > high amps and low volts rather than the other way around? It depends on what you are using it for. Low voltage isn't as dangerous. Low voltage requires big wires and is lossy. etc etc etc etc For cars, notice that 12V is now standard. Used to be you could find 6V automobile electrical systems. Go too low in voltage and it's a bad idea. (Also makes for dim headlights in practice....) >Does increased amps > lead to increased power in the same proportion as increased voltages leads to > increased power? Using Ohm's Law, yes. This characteristic is not useful by itself. >Why are car batteries touted as having more cranking amps, > isn't your amperage held constant by: > > V = I R V = 12, R = Constant Power = V^2/R > > What am I missing here? A brain perhaps? > Thanks in advance, -Mark [What you're missing is the background to analyse the problem. (Practical EE as opposed to mathematical EE)] V is not constant, I is not constant because R is not constant. A starter motor is only a long length of wire until it begins to turn. (It develops something called 'counter EMF' which tends to oppose current flow.) Battery output voltage tends to drop with increasing load current due to the internal resistance of the battery (I**2 R losses) The only reason why "cranking amps" is important is that it takes considerable power to budge a car engine. The more amperage available, the more power is available to the starter motor to crank over the engine. (Cranking amps may also give insight into cold temperature battery performance.) A practical 'motors 'n rotors ' class would make sense of all this. I didn't get mine in college. I got it in Naval Nuclear Power School in order to deal with BIIIIG !!! motors 'n rotors. :-) Ron Miller